\(\int \frac {1}{x^4 (a^2+2 a b x^2+b^2 x^4)} \, dx\) [491]

Optimal result

Integrand size = 24, antiderivative size = 68 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=-\frac {5}{6 a^2 x^3}+\frac {5 b}{2 a^3 x}+\frac {1}{2 a x^3 \left (a+b x^2\right )}+\frac {5 b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}} \]

[Out]

-5/6/a^2/x^3+5/2*b/a^3/x+1/2/a/x^3/(b*x^2+a)+5/2*b^(3/2)*arctan(x*b^(1/2)/a^(1/2))/a^(7/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {28, 296, 331, 211} \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {5 b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}}+\frac {5 b}{2 a^3 x}-\frac {5}{6 a^2 x^3}+\frac {1}{2 a x^3 \left (a+b x^2\right )} \]

[In]

Int[1/(x^4*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]

[Out]

-5/(6*a^2*x^3) + (5*b)/(2*a^3*x) + 1/(2*a*x^3*(a + b*x^2)) + (5*b^(3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7/2
))

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps \begin{align*} \text {integral}& = b^2 \int \frac {1}{x^4 \left (a b+b^2 x^2\right )^2} \, dx \\ & = \frac {1}{2 a x^3 \left (a+b x^2\right )}+\frac {(5 b) \int \frac {1}{x^4 \left (a b+b^2 x^2\right )} \, dx}{2 a} \\ & = -\frac {5}{6 a^2 x^3}+\frac {1}{2 a x^3 \left (a+b x^2\right )}-\frac {\left (5 b^2\right ) \int \frac {1}{x^2 \left (a b+b^2 x^2\right )} \, dx}{2 a^2} \\ & = -\frac {5}{6 a^2 x^3}+\frac {5 b}{2 a^3 x}+\frac {1}{2 a x^3 \left (a+b x^2\right )}+\frac {\left (5 b^3\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{2 a^3} \\ & = -\frac {5}{6 a^2 x^3}+\frac {5 b}{2 a^3 x}+\frac {1}{2 a x^3 \left (a+b x^2\right )}+\frac {5 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=-\frac {1}{3 a^2 x^3}+\frac {2 b}{a^3 x}+\frac {b^2 x}{2 a^3 \left (a+b x^2\right )}+\frac {5 b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}} \]

[In]

Integrate[1/(x^4*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]

[Out]

-1/3*1/(a^2*x^3) + (2*b)/(a^3*x) + (b^2*x)/(2*a^3*(a + b*x^2)) + (5*b^(3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^
(7/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.81

method	result	size
default	\(-\frac {1}{3 a^{2} x^{3}}+\frac {2 b}{a^{3} x}+\frac {b^{2} \left (\frac {x}{2 b \,x^{2}+2 a}+\frac {5 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}\)	\(55\)
risch	\(\frac {\frac {5 b^{2} x^{4}}{2 a^{3}}+\frac {5 b \,x^{2}}{3 a^{2}}-\frac {1}{3 a}}{x^{3} \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {-a b}\, b \ln \left (-b x -\sqrt {-a b}\right )}{4 a^{4}}-\frac {5 \sqrt {-a b}\, b \ln \left (-b x +\sqrt {-a b}\right )}{4 a^{4}}\)	\(91\)

[In]

int(1/x^4/(b^2*x^4+2*a*b*x^2+a^2),x,method=_RETURNVERBOSE)

[Out]

-1/3/a^2/x^3+2*b/a^3/x+b^2/a^3*(1/2*x/(b*x^2+a)+5/2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.53 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\left [\frac {30 \, b^{2} x^{4} + 20 \, a b x^{2} + 15 \, {\left (b^{2} x^{5} + a b x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) - 4 \, a^{2}}{12 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}, \frac {15 \, b^{2} x^{4} + 10 \, a b x^{2} + 15 \, {\left (b^{2} x^{5} + a b x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) - 2 \, a^{2}}{6 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}\right ] \]

[In]

integrate(1/x^4/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")

[Out]

[1/12*(30*b^2*x^4 + 20*a*b*x^2 + 15*(b^2*x^5 + a*b*x^3)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 +
a)) - 4*a^2)/(a^3*b*x^5 + a^4*x^3), 1/6*(15*b^2*x^4 + 10*a*b*x^2 + 15*(b^2*x^5 + a*b*x^3)*sqrt(b/a)*arctan(x*
sqrt(b/a)) - 2*a^2)/(a^3*b*x^5 + a^4*x^3)]

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.68 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=- \frac {5 \sqrt {- \frac {b^{3}}{a^{7}}} \log {\left (- \frac {a^{4} \sqrt {- \frac {b^{3}}{a^{7}}}}{b^{2}} + x \right )}}{4} + \frac {5 \sqrt {- \frac {b^{3}}{a^{7}}} \log {\left (\frac {a^{4} \sqrt {- \frac {b^{3}}{a^{7}}}}{b^{2}} + x \right )}}{4} + \frac {- 2 a^{2} + 10 a b x^{2} + 15 b^{2} x^{4}}{6 a^{4} x^{3} + 6 a^{3} b x^{5}} \]

[In]

integrate(1/x**4/(b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

-5*sqrt(-b**3/a**7)*log(-a**4*sqrt(-b**3/a**7)/b**2 + x)/4 + 5*sqrt(-b**3/a**7)*log(a**4*sqrt(-b**3/a**7)/b**2
+ x)/4 + (-2*a**2 + 10*a*b*x**2 + 15*b**2*x**4)/(6*a**4*x**3 + 6*a**3*b*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {15 \, b^{2} x^{4} + 10 \, a b x^{2} - 2 \, a^{2}}{6 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}} + \frac {5 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} \]

[In]

integrate(1/x^4/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")

[Out]

1/6*(15*b^2*x^4 + 10*a*b*x^2 - 2*a^2)/(a^3*b*x^5 + a^4*x^3) + 5/2*b^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {5 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} + \frac {b^{2} x}{2 \, {\left (b x^{2} + a\right )} a^{3}} + \frac {6 \, b x^{2} - a}{3 \, a^{3} x^{3}} \]

[In]

integrate(1/x^4/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")

[Out]

5/2*b^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) + 1/2*b^2*x/((b*x^2 + a)*a^3) + 1/3*(6*b*x^2 - a)/(a^3*x^3)

Mupad [B] (verification not implemented)

Time = 14.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {\frac {5\,b\,x^2}{3\,a^2}-\frac {1}{3\,a}+\frac {5\,b^2\,x^4}{2\,a^3}}{b\,x^5+a\,x^3}+\frac {5\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{7/2}} \]

[In]

int(1/(x^4*(a^2 + b^2*x^4 + 2*a*b*x^2)),x)

[Out]

((5*b*x^2)/(3*a^2) - 1/(3*a) + (5*b^2*x^4)/(2*a^3))/(a*x^3 + b*x^5) + (5*b^(3/2)*atan((b^(1/2)*x)/a^(1/2)))/(2
*a^(7/2))